Problem: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x + 2)^2 - 16$ $\text{lesser }x = $
$\begin{aligned} (x + 2)^2 - 16&= 0 \\\\ (x+2)^2&=16 \\\\ \sqrt{(x+2)^2}&=\sqrt{16} \end{aligned}$ $\begin{aligned} x+2&=\pm4 \\\\ x&=\pm4-2 \\ \phantom{(x + 2)^2 - 16}& \\ x=-6&\text{ or }x=2 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -6 \\\\ \text{greater } x &= 2 \end{aligned}$